Linear Differental Equations

First, we need to write the DE in standard form:

\( y^{'} - \frac{3}{t} y = \frac{\sin {\pi t}}{t^2} \)

Where \(P(t) = - \frac{3}{t} \)

The integrating factor is:

\( \mu (t) = e^ {\int P(t)} = e^{\int P(t)} = e^ {\int - \frac{3}{t} dt} = e^{-3 \ln t } = t^{-3} \)

\( \mu (t) = t^{-3} \)

First, write the DE in standard form:

\( y^{'} + \cfrac{4}{t} y = \cfrac{e^{-t}}{t^3} \)

Next, identify \(P(x)\) and find the integrating factor:

\( \mu (x) = e^ {\int P(x)} \)

\( \mu (t) = e^ {\int \cfrac{4}{t} dt} \)

\( \mu (t) = e^ {4 \ln t} \)

\( \mu (t) = t^4 \)

Next, multiply both sides of the DE by the integration factor:

\( y^{'} + \cfrac{4}{t} y = \cfrac{e^{-t}}{t^3} \)

\( t^4(y^{'} + \cfrac{4}{t} y) = t^4 \cfrac{e^{-t}}{t^3} \)

Replace the left hand side of the equation and simplify the right hand side:

\( \cfrac{d}{dt}(t^4 y) = t e^{-t} \)

Notice the left hand side, by chain rule is:

\( \cfrac{d}{dt}(t^4 y) = t^4 y^{'} + 4t^3y = t^4(y^{'} +\cfrac{4}{t}y) \)

Next, take the integral of both sides:

\( \int{ \cfrac{d}{dt}(t^4 y) dt} = \int{ t e^{-t} dt} \)

The left hand side integral and derivative cancel out. The right hand side can be solved using integration by parts:

\( u = t\)	\( u^{'} = 1\)
\( v = e^{-t}\)	\( \int v = -e^{-t}\)

\( \int{ t e^{-t} dt} = -te^{-t} - \int {-e^{-t} dt} \)

\( \int{ t e^{-t} dt} = -te^{-t} - e^{-t} + c\)

\( \int{ \cfrac{d}{dt}(t^4 y) dt} = \int{ t e^{-t} dt} \)

\( t^4 y = -te^{-t} - e^{-t} + C \)

\( y= \cfrac{-e^{-t}}{t^3} - \cfrac{e^{-t}}{t^4} + \cfrac{C}{t^4} \)

Now we can solve for the constant using \(y(-1) = 0 \):

\( y(-1) = 0 = \cfrac{-e^{1}}{-1} - \cfrac{e^{1}}{1} + \cfrac{C}{1} \)

\( 0 = C \)

\( y = \cfrac{-e^{-t}}{t^3} - \cfrac{e^{-t}}{t^4} \)

The solution to the DE is:

\( y = -(t+1)\cfrac{e^{-t}}{t^4} \)